The minimal growth of entire functions with given zeros along unbounded sets

Andrusyak, I. V.; Filevych, P. V.

doi:10.30970/ms.54.2.146-153

Cited by 4 publications

(8 citation statements)

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“…Many authors (see, for example, [1-5, 7, 8, 13, 15, 16]) investigated the opposite, in a certain sense, and more interesting question: how slow can the growth of a function f ∈ E (ζ) be with respect to n ζ (r) or N ζ (r)? In particular, the following two theorems that generalize two results of W. Bergweiler [8] were proved in [4].…”

Section: Introduction and Resultsmentioning

confidence: 82%

“…Note also that the proof of Theorem B given in [4] shows that the set F in this theorem can be expressed in the form F = ∞ k=1 (s k , r k ), where…”

Section: Introduction and Resultsmentioning

confidence: 96%

“…It is clear that if (2) is not satisfied for some positive function ψ on R, then log n ζ (r) in ( 1) can be replaced by ψ log n ζ (r) . According to Theorem B, the same replacement is impossible when (2) is satisfied.…”

Section: Introduction and Resultsmentioning

confidence: 99%

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Comparative growth of an entire function and the integrated counting function of its zeros

Andrusyak,

Filevych

2024

Carpathian Math. Publ.

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Let $(\zeta_n)$ be a sequence of complex numbers such that $\zeta_n\to\infty$ as $n\to\infty$, $N(r)$ be the integrated counting function of this sequence, and let $\alpha$ be a positive continuous and increasing to $+\infty$ function on $\mathbb{R}$ for which $\alpha(r)=o(\log (N(r)/\log r))$ as $r\to+\infty$. It is proved that for any set $E\subset(1,+\infty)$ satisfying $\int_{E}r^{\alpha(r)}dr=+\infty$, there exists an entire function $f$ whose zeros are precisely the $\zeta_n$, with multiplicities taken into account, such that the relation $$ \liminf_{r\in E,\ r\to+\infty}\frac{\log\log M(r)}{\log r\log (N(r)/\log r)}=0 $$ holds, where $M(r)$ is the maximum modulus of the function $f$. It is also shown that this relation is best possible in a certain sense.

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Section: Introduction and Resultsmentioning

confidence: 82%

“…Note also that the proof of Theorem B given in [4] shows that the set F in this theorem can be expressed in the form F = ∞ k=1 (s k , r k ), where…”

Section: Introduction and Resultsmentioning

confidence: 96%

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Comparative growth of an entire function and the integrated counting function of its zeros

Andrusyak,

Filevych

2024

Carpathian Math. Publ.

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“…At the end of the introductory part, we note that some other problems concerning comparisons of the growth of an entire function f to the distribution of its zeros were considered, in particular, in [3,4,5,6,7,8,9,10]. We also note that questions regarding the sizes of exceptional sets in various asymptotic relations between characteristics of entire functions were investigated, for example, in [12,13,14,15,16,17,18,19].…”

Section: Theorem C ([2]mentioning

confidence: 99%

“…Consider a function f ∈ E(ζ) and prove that this function satisfies ( 6) with a set F (f ) satisfying (7).…”

Section: Lemma 2 ([21]mentioning

confidence: 99%

Minimal growth of entire functions with prescribed zeros outside exceptional sets

2022

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Let $h$ be a positive continuous increasing to $+\infty$ function on $\mathbb{R}$. It is proved that for an arbitrary complex sequence $(\zeta_n)$ such that $0<|\zeta_1|\le|\zeta_2|\le\dots$ and $\zeta_n\to\infty$ as $n\to\infty$, there exists an entire function $f$ whose zeros are the $\zeta_n$, with multiplicities taken into account, for which$$\ln m_2(r,f)=o(N(r)),\quad r\notin E,\ r\to+\infty.$$with a set $E$ satisfying $\int_{E\cap(1,+\infty)}h(r)dr<+\infty$, if and only if $\ln h(r)=O(\ln r)$ as $r\to+\infty$.Here $N(r)$ is the integrated counting function of the sequence $(\zeta_n)$ and$$m_2(r,f)=\left(\frac{1}{2\pi}\int_0^{2\pi}|\ln|f(re^{i\theta})||^2d\theta\right)^{1/2}.$$

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Asymptotic estimates for entire functions of minimal growth with given zeros

Filevych

2024

Mat. Stud.

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Let $\zeta=(\zeta_n)$ be an arbitrary complex sequence such that $0<|\zeta_1|\le|\zeta_2|\le\dots$ and $\zeta_n\to\infty$ as $n\to\infty$, let $n_\zeta(r)$ and $N_\zeta(r)$ be the counting function and the integrated counting function of this sequence, respectively. By $\mathcal{E}_\zeta$ we denote the class of all entire functions whose zeros are precisely the $\zeta_n$, where a complex number that occurs $m$ times in the sequence $\zeta$ corresponds to a zero of multiplicity $m$. Suppose that $\Phi$ is a convex function on $\mathbb{R}$ such that $\Phi(\sigma)/\sigma\to+\infty$ as $\sigma\to+\infty$. It is proved that there exists an entire function $f\in\mathcal{E}_\zeta$ such that$$\varlimsup_{r\to+\infty}\frac{\ln\ln M_f(r)}{\Phi(\ln r)}\le\varlimsup_{r\to+\infty}\frac{\ln n_\zeta( r)}{\Phi(\ln r)},$$where $M_f(r)$ denotes the maximum modulus of the function $f$, and it is shown that the above inequality implies the inequality$$\varlimsup_{r\to+\infty}\frac{\ln\ln M_f(r)}{\Phi(\ln r)}\le\varlimsup_{r\to+\infty}\frac{\ln N_\zeta( r)}{\Phi(\ln r)}+\varlimsup_{\sigma\to+\infty}\frac{\ln\Phi'_+(\sigma)}{\Phi(\sigma)}.$$The formulated result is a consequence of the following more general statement: if the right-hand derivative $\Phi'_+$ of the function $\Phi$ assumes only integer values and $\sum_{n=1}^\infty e^{-\Phi(\ln|\zeta_n|)}<+\infty$, then there exists an entire function $f\in\mathcal{E}_\zeta$ such that $\ln M_f(r)=o(e^{\Phi(\ln r)})$ as $r \to+\infty$.

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The minimal growth of entire functions with given zeros along unbounded sets

Cited by 4 publications

References 1 publication

Comparative growth of an entire function and the integrated counting function of its zeros

Comparative growth of an entire function and the integrated counting function of its zeros

Minimal growth of entire functions with prescribed zeros outside exceptional sets

Asymptotic estimates for entire functions of minimal growth with given zeros

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